POJ 3281 Dining

题目链接

题意：n个牛。每一个牛有一些喜欢的食物和饮料。每种食物饮料仅仅有一个。问最大能匹配上多少仅仅牛每一个牛都能吃上喜欢的食物和喜欢的饮料

思路：最大流。建模源点到每一个食物连一条边，容量为1，每一个饮料向汇点连一条边容量为1，然后因为每一个牛有容量1。所以把牛进行拆点。然后食物连向牛的入点，牛的出点连向食物，跑一下最大流就可以

代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int MAXNODE = 505;
const int MAXEDGE = 100005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
	int u, v;
	Type cap, flow;
	Edge() {}
	Edge(int u, int v, Type cap, Type flow) {
		this->u = u;
		this->v = v;
		this->cap = cap;
		this->flow = flow;
	}
};

struct Dinic {
	int n, m, s, t;
	Edge edges[MAXEDGE];
	int first[MAXNODE];
	int next[MAXEDGE];
	bool vis[MAXNODE];
	Type d[MAXNODE];
	int cur[MAXNODE];
	vector<int> cut;

	void init(int n) {
		this->n = n;
		memset(first, -1, sizeof(first));
		m = 0;
	}
	void add_Edge(int u, int v, Type cap) {
		edges[m] = Edge(u, v, cap, 0);
		next[m] = first[u];
		first[u] = m++;
		edges[m] = Edge(v, u, 0, 0);
		next[m] = first[v];
		first[v] = m++;
	}

	bool bfs() {
		memset(vis, false, sizeof(vis));
		queue<int> Q;
		Q.push(s);
		d[s] = 0;
		vis[s] = true;
		while (!Q.empty()) {
			int u = Q.front(); Q.pop();
			for (int i = first[u]; i != -1; i = next[i]) {
				Edge& e = edges[i];
				if (!vis[e.v] && e.cap > e.flow) {
					vis[e.v] = true;
					d[e.v] = d[u] + 1;
					Q.push(e.v);
				}
			}
		}
		return vis[t];
	}

	Type dfs(int u, Type a) {
		if (u == t || a == 0) return a;
		Type flow = 0, f;
		for (int &i = cur[u]; i != -1; i = next[i]) {
			Edge& e = edges[i];
			if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {
				e.flow += f;
				edges[i^1].flow -= f;
				flow += f;
				a -= f;
				if (a == 0) break;
			}
		}
		return flow;
	}

	Type Maxflow(int s, int t) {
		this->s = s; this->t = t;
		Type flow = 0;
		while (bfs()) {
			for (int i = 0; i < n; i++)
				cur[i] = first[i];
			flow += dfs(s, INF);
		}
		return flow;
	}

	void MinCut() {
		cut.clear();
		for (int i = 0; i < m; i += 2) {
			if (vis[edges[i].u] && !vis[edges[i].v])
				cut.push_back(i);
		}
	}
} gao;

int f, d, n;

int main() {
	while (~scanf("%d%d%d", &n, &f, &d)) {
		int s = 0, t = f + d + n * 2 + 1;
		gao.init(t + 1);
		for (int i = 1; i <= f; i++)
			gao.add_Edge(s, i, 1);
		for (int i = 1; i <= d; i++)
			gao.add_Edge(f + i, t, 1);
		for (int i = 1; i <= n; i++) {
			gao.add_Edge(i + f + d, n + f + d + i, 1);
			int fn, dn;
			scanf("%d%d", &fn, &dn);
			int v;
			while (fn--) {
				scanf("%d", &v);
				gao.add_Edge(v, i + f + d, 1);
			}
			while (dn--) {
				scanf("%d", &v);
				gao.add_Edge(n + f + d + i, f + v, 1);
			}
		}
		printf("%d\n", gao.Maxflow(s, t));
	}
	return 0;
}